1
$\begingroup$

In the book "Rational Points on Elliptic Curves by J.Silverman and J.Tate" it is defined a representation $$ \rho_n:Gal(\mathbb{Q}(E[n])/\mathbb{Q})\longrightarrow GL_2(\mathbb{Z}/n\mathbb{Z})\hspace{0.2cm}\forall n\geq 2$$

with $\mathbb{Q}(E[n]):=\mathbb{Q}(x_1,y_1,...,x_{n^2-1},y_{n^2-1})$, where $E[n]=\{O,(x_1,y_1),...,(x_{n^2-1},y_{n^2-1})\}$

But in other refferences i saw $\rho_n:Gal(\overline{\mathbb{Q}}/\mathbb{Q})\longrightarrow GL_2(\mathbb{Z}/n\mathbb{Z})\hspace{0.2cm}\forall n\geq 2$

I know $\mathbb{Q}(E[n])\subset \overline{\mathbb{Q}}$ beacuse $\mathbb{Q}(E[n])/\mathbb{Q}$ is algebraic, in fact, it is Galois.

Which Galois group should i use?

$\endgroup$

1 Answer 1

2
$\begingroup$

The map $\operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q)\to\operatorname{GL}_2(\mathbb Z/n\mathbb Z)$ factors through $\operatorname{Gal}({\mathbb Q}(E[n])/\mathbb Q)\to\operatorname{GL}_2(\mathbb Z/n\mathbb Z)$. So it's not really a matter of which Galois group to use, it's a matter of what you're doing. If $n$ is fixed, it's often easier to use the finite Galois group; but if you're going to take a sequence of increasing values of $n$, e.g., $n=\ell^k$ with $k\to\infty$, then you're better off using the Galois group of $\overline{\mathbb Q}$ so that you don't have to keep switching groups.

$\endgroup$
2
  • $\begingroup$ Thank you very much!. I have a question about it. I saw $\overline{\mathbb{Q}}/\mathbb{Q}$ and $\mathbb{Q}(E[n])/\mathbb{Q}$ are Galois extension. Why is the Galois condition necessary in order to define representations? I think if $L/\mathbb{Q}$ is any extension and $\sigma:L\longrightarrow \overline{\mathbb{Q}}$ is a homomorphism of fields then $\sigma(x,y)=(\sigma(x),\sigma(y))$ define an action over $E[n]$ without Galois condition. $\endgroup$ Commented May 8, 2020 at 19:22
  • 1
    $\begingroup$ @danihelovick A (linear) representation, by definition, is a group homomorphism from a group $G$ to a linear group GL$(V)$. So you can't really call it a representation. OTOH, sure, $\sigma$ will define a group action, as long as $E[n]\subset E(L)$. But if not, then $\sigma$ does not give a well-defined map from $E[n]$ to itself. $\endgroup$ Commented May 8, 2020 at 20:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .